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Foundations+of+circuits 2

1. 1. chapter 22.1 TERMINOLOGY2.2 KIRCHHOFF’S LAWS2.3 CIRCUIT ANALYSIS: BASIC METHOD2.4 INTUITIVE METHOD OF CIRCUIT ANALYSIS: SERIES AND PARALLEL SIMPLIFICATION2.5 MORE CIRCUIT EXAMPLES2.6 DEPENDENT SOURCES AND THE CONTROL CONCEPT2.7 A FORMULATION SUITABLE FOR A COMPUTER SOLUTION *2.8 SUMMARY EXERCISES PROBLEMS
2. 2. resistive networks 2A simple electrical network made from a voltage source and four resistors isshown in Figure 2.1. This might be an abstract representation of some realelectrical network, or a model of some other physical system, for example, aheat flow problem in a house. We wish to develop systematic general methodsfor analyzing circuits such as this, so that circuits of arbitrary complexity canbe solved with dispatch. Solving or analyzing a circuit generally involves findingthe voltage across, and current through, each of the circuit elements. Systematicgeneral methods will also enable us to automate the solution techniques so thatcomputers can be used to analyze circuits. Later on in this chapter and in thenext, we will show how our problem formulation facilitates direct computeranalysis. To make the problem specific, suppose that we wish to find the currenti4 in Figure 2.1, given the values of the voltage source and the resistors. Ingeneral, we can resort to Maxwell’s Equations to solve the circuit. But thisapproach is really impractical. Instead, when circuits obey the lumped matterdiscipline, Maxwell’s Equations can be dramatically simplified into two algebraicrelationships stated as Kirchhoff’s voltage law (KVL) and Kirchhoff’s current law(KCL). This chapter introduces these algebraic relationships and then uses themto develop a systematic approach to solving circuits, thereby finding the currenti4 in our specific example. This chapter first reviews some terminology that will be useful in our discus-sions. We will then introduce Kirchhoff’s laws and work out some examplesto develop our facility with these laws. We will then introduce a systematicmethod for solving circuits based on Kirchhoff’s laws using a very simple, d a b + + V F I G U R E 2.1 Simple resistive − v4 i4 network. − c 53
3. 3. 54 CHAPTER TWO resistive networks illustrative circuit. We will then apply the same systematic method to solve more complicated examples, including the one shown in Figure 2.1. 2.1 T E R M I N O L O G Y Lumped circuit elements are the fundamental building blocks of electronic cir- cuits. Virtually all of our analyses will be conducted on circuits containing two-terminal elements; multi-terminal elements will be modeled using combi- nations of two-terminal elements. We have already seen several two-terminal elements such as resistors, voltage sources, and current sources. Electronic access to an element is made through its terminals. An electronic circuit is constructed by connecting together a collection of separate elements at their terminals, as shown in Figure 2.2. The junction points at which the terminals of two or more elements are connected are referred to as the nodes of a circuit. Similarly, the connections between the nodes are referred to as the edges or branches of a circuit. Note that each element in Figure 2.2 forms a single branch. Thus an element and a branch are the same for circuits comprising only two-terminal elements. Finally, circuit loops are defined to be closed paths through a circuit along its branches. Several nodes, branches, and loops are identified in Figure 2.2. In the circuit in Figure 2.2, there are 10 branches (and thus, 10 elements) and 6 nodes. As another example, a is a node in the circuit depicted in Figure 2.1 at which three branches meet. Similarly, b is a node at which two branches meet. ab and bc are examples of branches in the circuit. The circuit has five branches and four nodes. Since we assume that the interconnections between the elements in a circuit are perfect (i.e., the wires are ideal), then it is not necessary for a set of elements to be joined together at a single point in space for their interconnection to be considered a single node. An example of this is shown in Figure 2.3. While the four elements in the figure are connected together, their connection does not occur at a single point in space. Rather, it is a distributed connection. Nodes Loop ElementsF I G U R E 2.2 An arbitrary circuit. Branch
4. 4. 2.2 Kirchhoff’s Laws CHAPTER TWO 55 Elements B B F I G U R E 2.3 Distributed C A C interconnections of four circuit A D elements that nonetheless occur D Distributed node at a single node. Ideal wires - F I G U R E 2.4 Voltage and current i deﬁnitions illustrated on a branch in v Branch a circuit. current Branch + voltageNonetheless, because the interconnections are perfect, the connection can beconsidered to be a single node, as indicated in the figure. The primary signals within a circuit are its currents and voltages, which wedenote by the symbols i and v, respectively. We define a branch current as thecurrent along a branch of the circuit (see Figure 2.4), and a branch voltage as thepotential difference measured across a branch. Since elements and branches arethe same for circuits formed of two-terminal elements, the branch voltages andcurrents are the same as the corresponding terminal variables for the elementsforming the branches. Recall, as defined in Chapter 1, the terminal variables foran element are the voltage across and the current through the element. As an example, i4 is a branch current that flows through branch bc in thecircuit in Figure 2.1. Similarly, v4 is the branch voltage for the branch bc.2.2 K I R C H H O F F ’ S L A W SKirchhoff’s current law and Kirchhoff’s voltage law describe how lumped-parameter circuit elements couple at their terminals when they are assembledinto a circuit. KCL and KVL are themselves lumped-parameter simplificationsof Maxwell’s Equations. This section defines KCL and KVL and justifies thatthey are reasonable using intuitive arguments.1 These laws are employed incircuit analysis throughout this book.1. The interested reader can refer to Section A.2 in Appendix A for a derivation of Kirchhoff’s lawsfrom Maxwell’s Equations under the lumped matter discipline.
5. 5. 56 CHAPTER TWO resistive networks 2.2.1 K C L Let us start with Kirchhoff’s current law (KCL). KCL The current flowing out of any node in a circuit must equal the current flowing in. That is, the algebraic sum of all branch currents flowing into any node must be zero. Put another way, KCL states that the net current that flows into a node through some of its branches must flow out from that node through its remaining branches. Referring to Figure 2.5, if the currents through the three branches into node a are ia , ib , and ic , then KCL states that ia + ib + ic = 0. Similarly, the currents into node b must sum to zero. Accordingly, we must have −ib − i4 = 0. KCL has a simple intuitive justification. Referring to the closed box-like surface depicted in Figure 2.5, it is easy to see that the currents ia , ib , and ic must sum to zero, for otherwise, there would be a continuous charge buildup at node a. Thus, KCL is simply a statement of the conservation of charge. Let us now illustrate the different interpretations of KCL with the help of Figure 2.6. Which interpretation you use depends upon convenience and the specific circuit you are trying to analyze. Figure 2.6 shows a node joining N branches. Each of the branches contains some two-terminal element, the specifics of which are not relevant to our discussion. Note that all branch cur- rents are defined to be positive into the node. Since KCL states that no net a Sa ia ib b i1 iN d Sc Sb + V ic - i4 i2 ... iN-1 cF I G U R E 2.5 Currents into a node in the network. F I G U R E 2.6 A node at which N branches join.
6. 6. 2.2 Kirchhoff’s Laws CHAPTER TWO 57 i1 i1 i2 i2 ... ... F I G U R E 2.7 Two series- connected circuit elements.current can flow into a node, it follows for the node in Figure 2.6 that N in = 0. (2.1) n=1 Next, by negating Equation 2.1, KCL becomes N (−in ) = 0. (2.2) n=1Since −in is a current defined to be positive out from the node in Figure 2.6,this second form of KCL states that no net current can flow out from a node.Finally, Equation 2.1 can be rearranged to take the form M N in = (−in ), (2.3) n=1 n=M+1which demonstrates that the current flowing into a node through one set ofbranches must flow out from the node through the remaining branches. An important simplification of KCL focuses on the two series-connectedcircuit elements shown in Figure 2.7. Taking KCL to state that no net currentcan flow into a node, the application of KCL at the node between the twoelements yields i1 − i2 = 0 ⇒ i1 = i2 . (2.4) Node 1 i1This result is important because it shows that the branch currents passing i3through two series-connected elements must be the same. That is, there is i2nowhere for the current i1 to go as it enters the node connecting the two ele-ments except to exit that node as i2 . In fact, with multiple applications of KCL, i4 i5this observation is extendible to a longer string of series-connected elements. Node 3Such an extension would show that a common branch current passes through Node 2 Node 4a string of series-connected elements. i6e x a m p l e 2. 1 a m o r e g e n e r a l u s e o f k c l To illustrate F I G U R E 2.8 A circuit illustratingthe more general use of KCL consider the circuit in Figure 2.8, which has six branches a more general use of KCL.
7. 7. 58 CHAPTER TWO resistive networks connecting four nodes. Again, taking KCL to state that no net current can flow into a node, the application of KCL to the four nodes in the circuit yields Node 1: 0 = −i1 − i2 − i3 (2.5) Node 2: 0 = i1 + i4 − i6 (2.6) Node 3: 0 = i2 − i4 − i5 (2.7) Node 4: 0 = i3 + i5 + i6 . (2.8) Note that because each branch current flows into exactly one node and out from exactly one node, each branch current appears exactly once in Equations 2.5 through 2.8 posi- tively, and exactly once negatively. This would also be true if Equations 2.5 through 2.8i1 = 1 A i3 = 3 A were all written to state that no net current can flow out from a node. Such patterns can i2 often be used to spot errors. It is also because each branch current flows into exactly one node and out from exactly i4 i5 one node that summing Equations 2.5 through 2.8 yields 0 = 0. This in turn shows that the four KCL equations are dependent. In fact, a circuit with N nodes will have only N − 1 independent statements of KCL. Therefore, when fully analyzing a circuit it i6 is both necessary and sufficient to apply KCL to all but one node. If some of the branch currents in a circuit are known, then it is possible that KCL alone can be used to find other branch currents in the circuit. For example, consider the circuitF I G U R E 2.9 The circuit in in Figure 2.8 with i1 = 1 A and i3 = 3 A, as shown in Figure 2.9. Using Equation 2.5,Figure 2.8 with two branch currents namely KCL for Node 1, it can be seen that i2 = −4 A. This is all that can be learnednumerically deﬁned. from KCL alone given the information in Figure 2.9. But, if we further know that i5 = −2 A, for example, we can learn from KCL applied to the other nodes that i4 = −2 A and i6 = −1 A. 2A 12 A e x a m p l e 2 .2 u s i n g k c l t o d e t e r m i n e a n u n k n o w n 3A b r a n c h c u r r e n t Figure 2.10 shows five branches meeting at a node in i some circuit. As shown in the figure, four of the branch currents are given. Determine i. 6A By KCL, the sum of all the currents entering a node must equal the sum of all the currents exiting the node. In other words,F I G U R E 2.10 Five branches 2 A + 3 A + 6 A = 12 A + imeeting at a node. Thus, i = −1A.
8. 8. 2.2 Kirchhoff’s Laws CHAPTER TWO 59 i1 = 3Acos(ωt) Arbitrary Circuit F I G U R E 2.11 Node x in a circuit x pulled out for display. i3 i2 = 6Acos(ωt)e x a m p l e 2. 3 k c l a p p l i e d t o a n a r b i t r a r y n o d ei n a c i r c u i t Figure 2.11 shows an arbitrary circuit from which we havegrabbed a node x and pulled it out for display. The node is a junction point for threewires with currents i1 , i2 , and i3 . For the given values of i1 and i2 , determine the valueof i3 .By KCL, the sum of all currents entering a node must be 0. Thus, i1 + i2 − i3 = 0Note that i3 is negated in this equation because it is defined to be positive for a currentexiting the node. Thus i3 is the sum of i1 and i2 and is given by i3 = i1 + i2 = 3 cos(ωt) + 6 cos(ωt) = 9 cos(ωt) 2Ae x a m p l e 2. 4 e v e n m o r e k c l Figure 2.12 shows a node connect-ing three branches. Two of the branches have current sources that supply the currentsshown. Determine the value of i.By KCL, the sum of all the currents entering a node must equal 0. Thus i 2A+1A+i=0 1Aand i = −3 A. F I G U R E 2.12 Node connecting Finally, it is important to recognize that current sources can be used to three branches.construct circuits in which KCL is violated. Several examples of circuits con-structed from current sources in which KCL is violated at every node are shownin Figure 2.13. We will not be concerned with such circuits here for two reasons.First, if KCL does not hold at a node, then electric charge must accumulate atthat node. This is inconsistent with the constraint of the lumped matter disci-pline that dq/dt be zero. Second, if a circuit were actually built to violate KCL,something would ultimately give. For example, the current sources might cease
9. 9. 60 CHAPTER TWO resistive networks to function as ideal sources as they oppose one another. In any case, the behav- ior of the real circuit would not be well modeled by the type of circuit shown 1A 2A in Figure 2.13, and so there is no reason to study the latter. 2.2.2 K V L Let us now turn our attention to Kirchhoff’s voltage law (KVL). KVL is applied to circuit loops, that is, to interconnections of branches that form closed paths 1A 2A 3A through a circuit. In a manner analogous to KCL, Kirchhoff’s voltage law can be stated as: KVL The algebraic sum of the branch voltages around any closed path in a 2A network must be zero. Alternatively, it states that the voltage between two nodes is independent of the path along which it is accumulated. 1A 3A In Figure 2.14, the loop starting at node a, proceeding through nodes b and c, and returning to a, is a closed path. In other words, the closed loop defined by the three circuit branches a → b, b → c, and c → a in Figure 2.14 is a closed path.F I G U R E 2.13 Circuits that According to KVL, the sum of the branch voltages around this loop is zero.violate KCL. That is, vab + vbc + vca = 0 In other words, v1 + v4 + v3 = 0 where we have taken the positive sign for each voltage when going from the positive terminal to the negative terminal. It is important that we are consistent in how we assign polarities to voltages as we go around the loop. A helpful mnemonic for writing KVL equations is to assign the polarity to a given voltage in accordance with the first sign encountered when traversing that voltage around the loop. Like KCL, KVL has an intuitive justification as well. Recall that the def- inition of the voltage between a pair of nodes in a circuit is the potential + v1 - d a bF I G U R E 2.14 Voltages in a + + v3 - + v4 -closed loop in the network. V - i4 c
10. 10. 2.2 Kirchhoff’s Laws CHAPTER TWO 61 + vM –1 vM + F I G U R E 2.15 A loop containing N branches. + v1 vN +difference between the two nodes. The potential difference between two nodesis the sum of the potential differences for the set of branches along any pathbetween the two nodes. For a loop, the start and end nodes are one and thesame, and there cannot be a potential difference between a node and itself.Thus, since potential differences equate to voltages, the sum of branch volt-ages along a loop must equal zero. By the same reasoning, since the voltagebetween any pair of nodes must be unique, it must be independent of thepath along which branch voltages are added. Notice from the definition of avoltage that KVL is simply an expression of the principle of conservation ofenergy. The different interpretations of KVL are illustrated with the help ofFigure 2.15, which shows a loop containing N branches. Consider first theloop in Figure 2.15 in which all branch voltages decrease in the clockwise direc-tion. Since KVL states that the sum of the branch voltages around a loop iszero, it follows for the loop in Figure 2.15 that N vn = 0. (2.9) n=1 Note that in summing voltages along a loop we have adopted the con-vention proposed earlier: A positive branch voltage is added to the sum if thepath enters the positive end of a branch. Otherwise a negative branch voltage isadded to the sum. Therefore, to arrive at Equation 2.9, we have traversed theloop in the clockwise direction. Next, by negating Equation 2.9, KVL becomes N (−vn ) = 0. (2.10) n=1Since −vn is a voltage defined to be positive in the opposite direction, thissecond form of KVL shows that KVL holds whether it is applied along theclockwise or counterclockwise path around the loop.
11. 11. 62 CHAPTER TWO resistive networks vM–1 Node M vM + +F I G U R E 2.16 A loop containingN branches with some of thevoltage deﬁnitions reversed. v1 vN + Node 1 + e x a m p l e 2 .5 p a t h i n d e p e n d e n c e o f k v l Consider the loop in Figure 2.16 in which some of the voltage definitions are reversed for convenience. Applying KVL to this loop yields M−1 N M−1 N vn + (−vn ) = 0 ⇒ vn = vn . (2.11) n=1 n=M n=1 n=M The second equality in Equation 2.11 demonstrates that the voltage between two nodes is independent of the path along which it accumulated. In this case, the second equality shows that the voltage between Nodes 1 and M is the same whether accumu- lated along the path up the left side of the loop or the path up the right side of the loop. An important simplification of KVL focuses on the two parallel-connected circuit elements shown in Figure 2.17. Starting from the upper node and apply- ing KVL in the counterclockwise direction around the loop between the two circuit elements yields + + v1 v2 v1 − v2 = 0 ⇒ v1 = v2 . (2.12) This result is important because it shows that the voltages across two parallel- connected elements must be the same. In fact, with multiple applications of KVL, this observation is extendible to a longer string of parallel-connected elements. Such an extension would show that a common voltage appears across all parallel-connected elements in the string.F I G U R E 2.17 Two parallel- e x a m p l e 2 .6 a m o r e g e n e r a l u s e o f k v l To illustrateconnected circuit elements. the more general use of KVL consider the circuit in Figure 2.18, which has six branches connecting four nodes. Four paths along the loops through the circuit are also defined in the figure; note that the external loop, Loop 4, is distinct from the other three. Applying
12. 12. 2.2 Kirchhoff’s Laws CHAPTER TWO 63 Loop 4 + + 1 v + v3 – – v2 – Loop 1 Loop 2 + F I G U R E 2.18 A circuit having v4 + v 5 four nodes and six branches. – – Loop 3 + v – 6KVL to the four loops yields Loop 1: 0 = −v1 + v2 + v4 (2.13) Loop 2: 0 = −v2 + v3 − v5 (2.14) Loop 3: 0 = −v4 + v5 − v6 (2.15) Loop 4: 0 = v1 + v6 − v3 (2.16) Note that the paths along the loops have been defined so that each branch voltage istraversed positively around exactly one loop and negatively around exactly one loop.It is for this reason that each branch voltage appears exactly once in Equations 2.13through 2.16 positively, and exactly once negatively. As with the application of KCL,such patterns can often be used to spot errors.It is also because each branch voltage is traversed exactly once positively and once + v =3V +negatively that summing Equations 2.13 through 2.16 yields 0 = 0. This in turn shows v1 = 1 V + 3 – v2that the four KVL equations are dependent. In general, a circuit with N nodes and B – –branches will have B − N + 1 loops around which independent applications of KVL can v4+ + vbe made. Therefore, while analyzing a circuit it is necessary to apply KVL only to these 5 – –loops, which will in total, traverse each branch at least once in the process.If some of the branch voltages in a circuit are known, then it is possible that KVL +v –alone can be used to find other branch voltages in the circuit. For example, consider 6the circuit in Figure 2.18 with v1 = 1 V and v3 = 3 V, as shown in Figure 2.19. UsingEquation 2.16, namely KVL for Loop 4, it can be seen that v6 = 2 V. This is all that F I G U R E 2.19 The circuit incan be learned from KVL alone given the information in Figure 2.19. But, if we further Figure 2.18 with two branchknow that v2 = 2 V, for example, we can learn from KVL applied to the other loops voltages numerically deﬁned.that v4 = −1 V and v5 = 1 V.
13. 13. 64 CHAPTER TWO resistive networks Finally, it is important to recognize that voltage sources can be used to con- struct circuits in which KVL is violated. Several examples of circuits constructed 1V + – + 2V – from voltage sources in which KVL is violated around every loop are shown in Figure 2.20. As with circuits that violate KCL, we will not be concerned with circuits that violate KVL, for two reasons. First, if KVL does not hold around a loop, then magnetic flux linkage will accumulate through that loop.1V + – 2V + – + 3V – This is inconsistent with the constraint of the lumped matter discipline that d B /dt = 0 outside the elements. Second, if a circuit were actually built to vio- late KVL, something would ultimately give. For example, the voltage sources might cease to function as ideal sources as they oppose one another. Alterna- 2V tively, the loop inductance might begin to accumulate flux linkage, leading to –+ high currents that would damage the voltage sources or their interconnections. 1V + – – 3V + In any case, the behavior of the real circuit would not be well modeled by the type of circuit shown in Figure 2.20, and so there is no reason to study the latter.F I G U R E 2.20 Circuits that e x a m p l e 2 .7 v o l t a g e s o u r c e s i n s e r i e s Two 1.5-V volt-violate KVL. age sources are connected in series as shown in Figure 2.21. What is the voltage v at their terminals? To determine v, employ, for example, a counterclockwise application of KVL around the circuit, treating the port formed by the two terminals as an element having voltage v. In this case, 1.5 V + 1.5 V − v = 0, which has for its solution v = 3 V. + + 1.5 V - e x a m p l e 2 .8 k v l The voltages across two of the elements in the circuit in v Figure 2.22 are measured as shown. What are the voltages, v1 and v2 , across the other + two elements? 1.5 V - Since element #1 is connected in parallel with element #4, the voltages across them - must be the same. Thus, v1 = 5 V. Similarly, the voltage across the series connection of elements #2 and #3 must also be 5 V, so v2 = 3 V. This latter result can also be obtained through the counterclockwise application of KVL around the loop including elementsF I G U R E 2.21 The series #2, #3, and #4, for example. This yields, v2 + 2 V − 5 V = 0. Again, v2 = 3 V.connection of two 1.5-V batteries. e x a m p l e 2 .9 verifying kvl for a circuit Verify that the branch voltages shown in Figure 2.23 satisfy KVL. Summing the voltages in the loop e, d, a, b, e, we get −3 − 1 + 3 + 1 = 0. Similarly, summing the voltages in the loop e, f, c, b, e, we get +1 − (−2) − 4 + 1 = 0.
14. 14. 2.2 Kirchhoff’s Laws CHAPTER TWO 65 + 3 V - + 4 V - a b c + + + 1 V 1 V -2 V - + 3 V - - - d e f + + 1 V - v2 #2 + + + + - v1 #1 #4 5 V 2 V 1 V + - - - - 2 V #3 g -F I G U R E 2.22 A circuit with two measured and two F I G U R E 2.23 A circuit with element voltages as shown.unmeasured voltages.Finally, summing the voltages in the loop g, e, f, g, we get −2 + 1 + 1 = 0.KVL is satisfied since the sum of the voltages around each of the three circuit loopsis zero.e x a m p l e 2. 10 s u m m i n g v o l t a g e s a l o n g d i f f e r e n tp a t h s Next, given the branch voltages shown in Figure 2.23, determine the volt-age vga between the nodes g and a by summing the branch voltages along the path g, e,d, a. Then, show that vga is the same if path g, f, c, b, a is chosen.Summing the voltage increases along the path g, e, d, a, we get vga = −2 V − 3 V − 1 V = −6 V.Similarly, summing the voltage increases along the path g, f, c, b, a we get vga = −1 V − (−2 V) − 4 V − 3 V = −6 V.Clearly, both paths yield −6 V for vga .
15. 15. 66 CHAPTER TWO resistive networks Thus far, Chapters 1 and 2 have shown that the operation of a lumped system is described by two types of equations: equations that describe the behavior of its individual elements, or element laws (Chapter 1), and equations that describe how its elements interact when they are connected to form the system, or KCL and KVL (Chapter 2). For an electronic circuit, the element laws relate the branch currents to the branch voltages of the elements. The interactions between its elements are described by KCL and KVL, which are also expressed in terms of branch currents and voltages. Thus, branch currents and voltages become the fundamental signals within a lumped electronic circuit. 2.3 C I R C U I T A N A L Y S I S : B A S I C M E T H O D We are now ready to introduce a systematic method of solving circuits. It is framed in the context of a simple class of circuits, namely circuits containing only sources and linear resistors. Many of the important analysis issues can be understood through the study of these circuits. Solving a circuit involves determining all the branch currents and branch voltages in the circuit. In practice, some currents or voltages may be more important than others, but we will not make that distinction yet. Before we return to the specific problem of analyzing the electrical network shown in Figure 2.1, let us first develop the systematic method using a few simpler circuits and build up our insight into the technique. We saw previously that under the lumped matter discipline, Maxwell’s Equations reduce to the basic element laws and the algebraic KVL and KCL. Accordingly, a systematic solution of the network involves the assembly and subsequent joint solution of two sets of equations. The first set of equations comprise the constituent relations for the individual elements in the network. The second set of equations results from the application of Kirchhoff’s current and voltage laws. This basic method of circuit analysis, also called the KVL and KCL method or the fundamental method, is outlined by the following steps: 1. Define each branch current and voltage in the circuit in a consistent manner. The polarities of these definitions can be arbitrary from one branch to the next. However, for any given branch, follow the associated variables convention (see Section 1.5.3 in Chapter 1). In other words, the branch current should be defined as positive into the positive voltage terminal of the branch. By following the associated variables, element laws can be applied consistently, and the solutions will follow a much clearer pattern. 2. Assemble the element laws for the elements. These element laws will specify either the branch current or branch voltage in the case of an independent source, or specify the relation between the branch current
16. 16. 2.3 Circuit Analysis: Basic Method CHAPTER TWO 67 and voltage in the case of a resistor. Examples of these element laws were presented in Section 1.6.3. Apply Kirchhoff’s current and voltage laws as discussed in Section 2.2.4. Jointly solve the equations assembled in Steps 2 and 3 for the branch variables defined in Step 1. The remainder of this chapter is devoted to circuit analysis examples thatrigorously follow these steps. Once the two sets of equations are assembled, which is a relatively easytask, the analysis of a circuit essentially becomes a problem of mathematics, asindicated by Step 4. That is, the equations assembled earlier must be combinedand used to solve for the branch currents and voltages of interest. However,because there is more than one way to approach this problem, our study ofcircuit analysis does not end with the direct approach outlined here. Consider-able time can be saved, and considerable insight can be gained, by approachingcircuit analyses in different ways. These gains are important subjects of thisand future chapters.2.3.1 S I N G L E - R E S I S T O R C I R C U I T STo illustrate our basic approach to circuit analysis, consider the simple circuitshown in Figure 2.24. The circuit has one independent source and one resistor, I Rand so has two branches, each with a current and a voltage. The goal of ourcircuit analysis is to find these branch variables. Step 1 in the analysis is to label the branch variables. We do so in Figure 2.25.Since there are two branches, there are two sets of variables. Notice that thebranch variables for the current source branch and for the resistor branch each F I G U R E 2.24 A circuit with only one independent current sourcefollow the associated variables convention. and one resistor. Now, we proceed with Steps 2 through 4: assemble the element laws,apply KCL and KVL, and then simultaneously solve the two sets of equationsto complete the analysis. i1 i2 The circuit has two elements. Following Step 2 we write the two elementlaws for these elements as + + I v1 v2 R i1 = −I, (2.17) - - v2 = Ri2 , (2.18) F I G U R E 2.25 Assignment ofrespectively. Here, v1 , i1 , v2 , and i2 are the branch variables. Note the dis- branch variables.tinction between the branch variable i1 and the source amplitude I. Here, theindependent source amplitude I is assumed to be known. Next, following Step 3, we apply KCL and KVL to the circuit. Since thecircuit has two nodes, it is appropriate to write KCL for one node, as discussed
17. 17. 68 CHAPTER TWO resistive networks in Section 2.2.1. The application of KCL at either node yields i1 + i2 = 0. (2.19) The circuit also has two branches that form one loop. So, following the discus- sion in Section 2.2.2 it is appropriate to write KVL for one loop. Starting at the upper node and traversing the loop in a clockwise manner, the application of KVL yields v2 − v1 = 0. (2.20) Notice we have used our mnemonic discussed in Section 2.2.2 for writing KVL equations. For example, in Equation 2.20, we have assigned a + polarity to v2 since we first encounter the + sign when traversing the branch with variable v2 . Similarly, we have assigned a − polarity to v1 since we first encounter the − sign when traversing the v1 branch. Finally, following Step 4, we combine Equations 2.17 through 2.20 and solve jointly to determine all four branch variables in Figure 2.25. This yields −i1 = i2 = I, (2.21) v1 = v2 = RI, (2.22) and completes the analysis of the circuit in Figure 2.25. i1 i2 + + e x a m p l e 2 .11 s i n g l e - r e s i s t o r c i r c u i t w i t h o n e + v1 v2 i n d e p e n d e n t v o l t a g e s o u r c e Now consider another simple V R - circuit shown in Figure 2.26. This circuit can be analyzed in an identical manner. It - - too has two elements, namely a voltage source and a resistor. Figure 2.26 already shows the definitions of branch variables, and so accomplishes Step 1. Next, following Step 2 we write the element laws for these elements asF I G U R E 2.26 A circuit with onlyone independent voltage source v1 = V (2.23)and one resistor. v2 = Ri2 , (2.24) respectively. Here, the independent source amplitude V is assumed to be known. Next, following Step 3, we apply KCL and KVL to the circuit. Since the circuit has two nodes, it is again appropriate to write KCL for one node. The application of KCL at either node yields i1 + i2 = 0. (2.25) The circuit also has two branches that form one loop, so it is again appropriate to write KVL for one loop. The application of KVL around the one loop in either direction yields v1 = v2 . (2.26)
18. 18. 2.3 Circuit Analysis: Basic Method CHAPTER TWO 69Finally, following Step 4, we combine Equations 2.23 through 2.26 to determine all fourbranch variables in Figure 2.26. This yields V −i1 = i2 = , (2.27) R v1 = v2 = V, (2.28)and completes the analysis of the circuit in Figure 2.26. For the circuit in Figure 2.25, there are four equations to solve for fourunknown branch variables. In general, a circuit having B branches will have 2Bunknown branch variables: B branch currents and B branch voltages. To findthese variables, 2B independent equations are required, B of which will comefrom element laws, and B of which will come from the application of KVL andKCL. Moreover, if the circuit has N nodes, then N−1 equations will come fromthe application of KCL and B − N + 1 equations will come from the applicationof KVL. While the two examples of circuit analysis presented here are admittedlyvery simple, they nonetheless illustrate the basic steps of circuit analysis: labelthe branch variables, assemble the element laws, apply KCL and KVL, andsolve the resulting equations for the branch variables of interest. While we willnot always follow these steps explicitly and in exactly the same order in futurechapters, it is important to know that we will nonetheless process exactly thesame information. It is also important to realize that the physical results of the analysis of thecircuit in Figure 2.25, and of any other circuit for that matter, cannot dependon the polarities of the definitions of the branch variables. We will work anexample to illustrate this point.e x a m p l e 2. 12 p o l a r i t i e s of branch variables i1Consider the analysis of the circuit in Figure 2.27, which is physically the same as thecircuit in Figure 2.25. The only difference in the two figures is the reversal of the polari- + -ties of i2 and v2 . The circuit in Figure 2.27 circuit has the same two elements, and theirelement laws are still I v1 v2 R - + i1 = −I (2.29) v2 = Ri2 . (2.30) i2Note that the polarity reversal of i2 and v2 has not changed the element law for the F I G U R E 2.27 A circuit similar toresistor from Equation 2.18 because the element law for a linear resistor is symmetric the one shown in Figure 2.24.when the terminal variables are defined according to the associated variables convention.The circuit also has the same two nodes and the same loop. The application of KCL at
19. 19. 70 CHAPTER TWO resistive networks either node now yields i1 − i2 = 0, (2.31) and the application of KVL around the loop now yields v1 + v2 = 0. (2.32) Note that Equations 2.31 and 2.32 differ from Equations 2.19 and 2.20 because of the polarity reversal of i2 and v2 . Finally, combining Equations 2.29 through 2.32 yields −i1 = −i2 = I (2.33) v1 = −v2 = RI, (2.34) which completes the analysis of the circuit in Figure 2.27. Now compare Equations 2.33 and 2.34 to Equations 2.21 and 2.22. The important observation here is that they are the same except for the polarity reversal of the solutions for i2 and v2 . This must be the case because the circuits in Figures 2.25 and 2.27 are physically the same, and so their branch variables must also be physically the same. Since we have chosen to define two of these branch variables with different polarities in the two figures, the signs of their values must differ accordingly so that they describe the same physical branch current and voltage. 2.3.2 Q U I C K I N T U I T I V E A N A L Y S I S O F SINGLE-RESISTOR CIRCUITS Before moving on to more complex circuits, it is worthwhile to analyze the circuit in Figure 2.25 in a more intuitive and efficient manner. Here, the element law for the current source directly states that i1 = −I. Next, the application of KCL to either node reveals that i2 = −i1 = I. In other words, the current from the source flows entirely through the resistor. Next, from the element law for the resistor, it follows that v2 = Ri2 = RI. Finally the application of KVL to the one loop yields v1 = v2 = RI to complete the analysis. e x a m p l e 2 .13 q u i c k intuitive analysis of a s i n g l e - r e s i s t o r c i r c u i t This example considers the circuit in Figure 2.26. Here, the element law for the voltage source directly states that v1 = V. Next, the application of KVL around the one loop reveals that v2 = v1 = V. In other words, the voltage from the source is applied directly across the resistor. Next, from the element law for the resistor, it follows that i2 = v2 /R = V/R. Finally, the application of KCL to either node yields i1 = −i2 = −V/R to complete the analysis. Notice that we had made use of a similar intuitive analysis in solving our battery and lightbulb example in Chapter 1. The important message here is that it is not necessary to first assemble all the circuit equations, and then solve them all at once. Rather, using a little
20. 20. 2.3 Circuit Analysis: Basic Method CHAPTER TWO 71intuition, it is likely to be much faster to approach the analysis in a differentmanner. We will have more to say about this in Section 2.4 and in Chapter 3.2.3.3 E N E R G Y C O N S E R V A T I O NOnce the branch variables of a circuit have been determined, it is possible toexamine the flow of energy through the circuit. This is often a very importantpart of circuit analysis. Among other things, such an examination should showthat energy is conserved in the circuit. This is the case for the circuits in Fig-ures 2.25 and 2.26. Using Equations 2.21 and 2.22 we see that the power intothe current source in Figure 2.25 is i1 v1 = −RI2 (2.35)and that the power into the resistor is i2 v2 = RI2 . (2.36)The negative sign in Equation 2.35 indicates that the current source actuallysupplies power. Similarly, using Equations 2.27 and 2.28 we see that the power into thevoltage source in Figure 2.26 is V2 i1 v1 = − (2.37) Rand that the power into the resistor is V2 i2 v2 = . (2.38) RIn both cases, the power generated by the source is equal to the power dissipatedin the resistor. Thus, energy is conserved in both circuits. Conservation of energy is itself an extremely powerful method for obtain-ing many types of results in circuits. It is particularly useful in dealing withcomplicated circuits that contain energy storage elements such as inductorsand capacitors that we will introduce in later chapters. Energy methods canoften allow us to obtain powerful results without a lot of mathematical grunge.We will use two energy-based approaches in this book.
21. 21. 72 CHAPTER TWO resistive networks 1. One energy approach equates the energy supplied by a set of elements in a circuit to the energy absorbed by the remaining set of elements in a circuit. Usually, this method involves equating the power generated by the devices in a circuit to the power dissipated in the circuit. 2. Another energy approach equates the total amount of energy in a system at two different points in time (assuming that there are no dissipative elements in the circuit). We will illustrate the use of the first method using a few examples in this section, and Section 9.5 in Chapter 9 will highlight examples using the second method. i e x a m p l e 2 .14 e n e r g y c o n s e r v a t i o n Determine the value + of v in the circuit in Figure 2.28 using the method of energy conservation. 2 mA v 1 kΩ We will show that the mathematical grunge of the basic method can be eliminated using - the energy method and some intuition. In Figure 2.28, the current source maintains a current i = 0.002 A through the circuit. To determine v, we equate the power supplied by the source to the power dissipated by the resistor. Since the current source and theF I G U R E 2.28 Energy resistor share terminals, the voltage v appears across the current source as well. Thus,conservation example. the power into the source is given by v × (−0.002) = −0.002v. In other words, the power supplied by the source is 0.002v. Next, the power into the resistor is given by v2 = 0.001v2 . 1k Finally, equating the power supplied by the source to the power dissipated by the resistor, we have 0.002v = 0.001v2 . In other words, v = 0.5 V. i = 3 mA? + + e x a m p l e 2 .15 u s i n g a n e n e r g y - b a s e d a p p r o a c h t o 3V v 1 kΩ - - v e r i f y a r e s u l t A student applies the basic method to the circuit in Figure 2.29 and obtains i = 3 mA. Determine whether this answer is correct by using the method of energy conservation.F I G U R E 2.29 Another energy By energy conservation, the power supplied by the source must be equal to the powerconservation example. dissipated by the resistor. Using the value of the current obtained by the student, the
22. 22. 2.3 Circuit Analysis: Basic Method CHAPTER TWO 73energy dissipated by the resistor is given by i2 × 1K = 9 mW.The energy into the voltage source is given by 3 V × 3 mA = 9 mW.In other words, the energy supplied by the source is given by −9 mW. Clearly the R1 + -energy supplied by the source is not equal to the energy dissipated by the resistor, and +so i = 3 mA is incorrect. Notice that if we reverse the polarity of i, energy will be v2 (a) R2conserved. Thus, i = −3 mA is the correct answer. -2.3.4 V O L T A G E A N D C U R R E N T D I V I D E R S R1We will now tackle several circuits called dividers that are slightly more complex + V (b)than the simplest single-loop, two-node, two-element circuits of the previous - + R2 v2section. These circuits will comprise a single loop and three or more elements, -or two nodes and three or more elements. Dividers produce fractions of inputcurrents or voltages and will be encountered often in subsequent chapters. For F I G U R E 2.30 Voltage-dividerthe moment, however, they are good examples on which to practice circuit circuit.analysis, and we can use them to gain important insight into circuit behavior.Voltage Dividers i1A voltage divider is an isolated loop that contains two or more resistors and a +voltage source in series. A physical voltage divider circuit is illustrated pictorially v1 R1in Figure 2.30a. We have connected two resistors in series, and connected the - -pair by some wires to a battery. Such a circuit is useful if we wish to obtain some +arbitrary fraction, say 10%, of the battery voltage at the terminals marked v2 . V v0To find the relation between v2 and the battery voltage and resistor values, we - i2 + +draw the circuit in schematic form, as shown in Figure 2.30b. We then follow v2the basic four-step method outlined in Section 2.3 to solve the circuit. R2 -1. The circuit has three elements, or branches, and hence it will have six i0 branch variables. Figure 2.31 shows one possible assignment of branch variables. To find these branch variables, we again assemble the element F I G U R E 2.31 Assignment of laws and the appropriate applications of KCL and KVL, and then branch variables to the voltage simultaneously solve the resulting equations. divider.2. The three element laws are v0 = −V (2.39)
23. 23. 74 CHAPTER TWO resistive networks v 1 = R1 i 1 (2.40) v2 = R2 i2 . (2.41) 3. Next, we apply KCL and KVL. The application of KCL to the two upper nodes yields i0 = i1 (2.42) i1 = i2 (2.43) and the application of KVL to the one loop yields v0 + v1 + v2 = 0. (2.44) 4. Finally, Equations 2.39 through 2.44 can be solved for the six unknown branch variables. This yields 1 i0 = i1 = i2 = V (2.45) R1 + R 2 and v0 = −V (2.46) R1 v1 = V (2.47) R1 + R 2 R2 v2 = V. (2.48) R1 + R 2 This completes the analysis of the two-resistor voltage divider. From the results of this analysis it should be apparent why the circuit in Figure 2.31 is called a voltage divider. Notice that v2 is some fraction (specifically, R2 /(R1 + R2 )) of the source voltage V, as desired. The fraction is the ratio of the resistance about which the voltage is measured and the sum of the resistances. By adjusting the relative values of R1 and R2 we can make this fraction adjust anywhere from 0 to 1. If, for example, we wish v2 to be one-tenth of V, as suggested at the start of this example, then R1 should be nine times as big as R2 . Notice also that v1 + v2 = V, and that the two resistors divide the voltage V in proportion to their resistances since v1 /v2 = R1 /R2 . For example, if R1 is twice R2 then v1 is twice v2 .
24. 24. 2.3 Circuit Analysis: Basic Method CHAPTER TWO 75 The voltage-divider relationship in terms of conductance can be foundfrom Equation 2.48 by substituting the conductances in place of the resistances: 1/G2 v2 = V (2.49) 1/G1 + 1/G2 G1 = V. (2.50) G1 + G 2Hence the voltage-divider relations expressed in terms of conductances involvethe conductance opposite the desired voltage, divided by the sum of the twoconductances. The simple circuit topology of Figure 2.30 is so common that the voltage-divider relation given by Equation 2.48 will become a primitive in our circuitvocabulary. It is helpful to build up a set of such primitives, which are reallysolved simple cases, to speed up circuit analysis, and to facilitate intuition.A simple mnemonic: For the voltage v2 , take the resistance associated with v2divided by the sum of the two resistances, multiplied by the voltage applied tothe pair.e x a m p l e 2. 16 v o l t a g e d i v i d e r A voltage divider circuit such asthat in Figure 2.30 has V = 10 V and R2 = 1 k . Choose R1 such that v2 is 10% of V.By the voltage divider relation of Equation 2.48, we have R2 v2 = V. R1 + R2For v2 to be 10% of V we must have v2 R2 = 0.1 = . V R1 + R2For R2 = 1 k , we must choose R1 such that 1k 0.1 = R1 + 1 kor R1 = 9 k .e x a m p l e 2. 17 t e m p e r a t u r e v a r i a t i o n Consider the cir-cuit in Figure 2.31 in which V = 5 V, R1 = 103 , and R2 = 103 (1 + T/(500 ◦ C)) ,where T is the temperature of the second resistor. Over what range does v2 vary if Tvaries over the range −100 ◦ C ≤ T ≤ 100 ◦ C?
25. 25. 76 CHAPTER TWO resistive networks Given the temperature range, R2 varies over the range: 0.8 × 103 ≤ R2 ≤ 1.2 × 103 . Therefore, following Equation 2.48, 2.2 V ≤ v2 ≤ 2.7 V, with the higher voltage occurring at the higher temperature. Having determined its branch variables we can now examine the flow of energy through the two-resistor voltage divider. Using Equations 2.45 through 2.48 we see that the power into the source is V2 i 0 v0 = − (2.51) R1 + R 2 and that the power into each resistor is R1 V2 i1 v1 = (2.52) (R1 + R2 )2 R2 V2 i2 v2 = . (2.53) (R1 + R2 )2 Since the power into the voltage source is the opposite of the total power into the two resistors, energy is conserved in the two-resistor voltage divider. That is, the power generated by the voltage source is exactly dissipated in the two resistors. Resistors in Series R1 R2 In electronic circuits one often finds resistors connected in series, as shown in Figures 2.31 and 2.32. For example, in our lightbulb example of Chapter 1, suppose the wire had a nonzero resistance, then the current through the wireF I G U R E 2.32 Resistors in would be related to the value of several resistances including those ofseries. the wires and the bulb in series. Our lumped circuit abstraction and the resulting Kirchhoff’s laws allow us to calculate the equivalent resistance of such combinations using simple algebra. Specifically, the analysis of the voltage divider shows that two resistors in series act as a single resistor having a resistance RS equal to the sum of the two individual resistances R1 and R2 . In other words, series resistances add. RS = R1 + R2 (2.54) To see this, observe that the voltage source in Figure 2.31 applies the volt- age V to two series resistors R1 and R2 , and that from Equation 2.43 these resistors respond with the common current i1 = i2 through their branches. Further, observe from Equation 2.45 that this common current, i = i1 = i2 ,
26. 26. 2.3 Circuit Analysis: Basic Method CHAPTER TWO 77is linearly proportional to the voltage from the source. Specifically, fromEquation 2.45, the common current is given by 1 i= V. (2.55) R1 + R 2 By comparing Equation 2.55 to Equation 1.4, we conclude that for tworesistors in series, the equivalent resistance of the pair when viewed from theirouter terminals is the sum of the individual resistance values. Specifically, ifRS is the resistance of the series resistor pair, then, from Equation 2.55, wefind that V RS = = R1 + R2 . (2.56) i This is consistent with the physical derivation of resistance in Equation 1.6since placing resistors in series essentially increases their combined length. By substituting their conductances, we can also obtain the equivalentconductance of a pair of conductances in series as 1 1 1 = + . (2.57) GS G1 G2Simplifying, G 1 G2 GS = . (2.58) G1 + G 2 As shown in the ensuing example, we can generalize our result for i1two series resistors to N resistors in series as: + RS = R1 + R2 + R3 + · · · RN . (2.59) v1 R1 - - Remember this result as another common circuit primitive. + V v0 - ... +e x a m p l e 2. 18 a n N-resistor voltage divider Nowconsider a more general voltage divider having N resistors, as shown in Figure 2.33. iNIt can be analyzed in the same manner as the two-resistor voltage divider. The only +difference is that there are now more unknowns to find, and hence more equations to vN RNwork with. To begin, suppose we assign the branch variables as shown in Figure 2.33. -The element laws are i0 v0 = −V (2.60) F I G U R E 2.33 A voltage divider vn = Rn in , 1 ≤ n ≤ N. (2.61) with N resistors.
27. 27. 78 CHAPTER TWO resistive networks Next, the application of KCL to the N − 1 upper nodes yields in = in−1 , 1 ≤ n ≤ N (2.62) and the application of KVL to the one loop yields v0 + v1 + · · · vN = 0. (2.63) Finally, Equations 2.60 through 2.63 can be solved to yield 1 in = V, 0 ≤ n ≤ N (2.64) R1 + R2 + · · · RN v0 = −V (2.65) Rn vn = V, 1 ≤ n ≤ N. (2.66) R1 + R2 + · · · RN This completes the analysis. As was the case for the two-resistor voltage divider, the preceding analysis shows that series resistors divide voltage in proportion to their resistances. This follows from the Rn in the numerator of the right-hand side of Equation 2.66. Additionally, the analysis again shows that series resistances add. To see this, let RS be the equivalent resistance of the N series resistors. Then, from Equation 2.64 we see that V RS = = R1 + R2 + · · · RN . (2.67) in This result is summarized in Figure 2.34. Finally, the two voltage-divider examples illustrate an important point, namely that series elements all carry the same branch current because the terminals from these elements are connected end-to-end without connection to additional branches through which the current can divert. This results in the KCL seen in Equations 2.42, 2.43, and 2.62, which state the equivalence of the branch currents. R1F I G U R E 2.34 The equivalence R2 RS = R1 + R2 + ... + RN ...of series resistors. RN
28. 28. 2.3 Circuit Analysis: Basic Method CHAPTER TWO 79e x a m p l e 2. 19 v o l t a g e - d i v i d e r c i r c u i t Determine v1 andv2 for the voltage-divider circuit in Figure 2.35 with R1 = 10 , R2 = 20 , andv(t) = 3 V using (a) the basic method and (b) the results from voltage dividers.(a) Let us first analyze the circuit using the basic method. 1. Assign variables as in Figure 2.36. 2. Write the constituent relations v0 = 3 V (2.68) v1 = 10i1 (2.69) v2 = 20i2 . (2.70) 3. Write KCL i1 − i2 = 0. (2.71) 4. Write KVL −v0 + v1 + v2 = 0. (2.72)Now eliminate i1 and i2 from Equations 2.69, 2.70, and 2.71, to obtain v2 v1 = . (2.73) 2Substituting this result and v0 = 3 V into Equation 2.72, we obtain v2 −3 V + + v2 = 0. (2.74) 2 i1 10 Ω + + - i2 v1 v1 R1 - + + + + v0 v2 Signal + 3V 20 Ω v(t) - generator R2 v2 - - - -F I G U R E 2.35 Voltage-divider circuit. F I G U R E 2.36 Voltage divider with variables assigned.
29. 29. 80 CHAPTER TWO resistive networks Hence, 2 v2 = 3V=2V (2.75) 3 and from Equation 2.73, v1 = 1 V. (b) Using the voltage-divider relation, we can write by inspection the value v2 as a function of the source voltage as follows: 20 v2 = 3 V = 2 V. 10 + 20 Similarly, 10 v1 = 3 V = 1 V. 10 + 20 Current Dividers i0 i2 A current divider is a circuit with two nodes joining two or more parallel resistors and a current source. Two current dividers are shown in Figures 2.37 and 2.38, + + + i1 the first with two resistors and the second with N resistors. In these circuits, I v0 v1 R1 v2 R2 the resistors share, or divide, the current from the source in proportion to their - - - conductances. It turns out that the equations for voltage dividers comprising voltages and resistances, and those for current dividers comprising currents and conductances, are very similar. Therefore, to highlight the duality betweenF I G U R E 2.37 A current divider these two types of circuits, we will attempt to mirror the steps from our voltagewith two resistors. divider discussion. Consider the two-resistor current divider shown in Figure 2.37. It has i0 three elements, or branches, and hence six unknown branch variables. To find ... these branch variables we again assemble the element laws and the appropriate + + i + iN 1 applications of KCL and KVL, and then simultaneously solve the resultingI v0 v1 R vN RN 1 equations. First, the three element laws are - - - ... i0 = −I (2.76)F I G U R E 2.38 A current divider v1 = R1 i1 (2.77)with N resistors. v2 = R2 i2 . (2.78) Next, the application of KCL to either node yields i0 + i1 + i2 = 0 (2.79) and the application of KVL to the two internal loops yields v0 = v1 (2.80) v1 = v2 . (2.81)